Question

6) If the percentage
error in calculating the
radius of a sphere
is 2%. What will be
the percentage error in calculating the volume​

Answer 1

Explanation:

The error in the measurement of the radius of a sphere is 2%. What will be the error in the calculation of its volume?

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The volume V of a sphere of radius r is

V=43πr3

If the error in measurement of the radius is ±2% , then it means that the actual radius r′ is such that

0.98r<r′<1.02r

This means that the actual volume V′ of the sphere is such that

43π(0.98r)3<V′<43π(1.02r)3

⟹(0.98)343πr3<V′<(1.02)343πr3

⟹0.941192V<V′<1.061208V

⟹(1−5.8808100)V<V′<(1+6.1208100)V

This means that the error in calculation of the volume of the sphere is between −5.8808% to 6.1208% .